StrmCkr wrote:we know that 3 is a used number, this implies that the # 5 can only be determined / is dependent on the # 2. Then i take this knowledge back into the inital equation.
(3+2)+2 and discover that the "2" becomes a repeating number in this equation creating a false solution. (thanks to the rules of the puzzle #'s cannot repeat).
there for
(5+2) ≠ 7
and this leaves
7| (4+2+1) as the only valid solution remaining and the only unknown # in this solution is 2
A = 2
This is the bit bothering me... I guess you're manipulating the candidate list of cell A (r4c7)... Obviously the sum of candidates in A must be 7 (Simple Sudoku determined that the candidate list must be {2,5})...
So, what you did, first you determined that the list must be either {2,5} or {1,2,4}, but for some reason you thought that {2,5} must be further broken up (
WHY? ) And must be either {2,{1,4}} or {2,{2,3}}, which were both impossible... But what about the original
{2,5}!Back to the start...
A=(45)-(9+8+7+6+4+3+1)... Here you're eliminating 7 different candidates out of a total of 9 possibilities... That shows you must have 2 possible condidates left, which sums to 7: {1,6}, {2,5} or {3,4}. But {1,6} and {3,4} all appeared on the cells which
A could see directly, therefore the only possibility is
{2,5}.
BUT you cannot eliminate 5 from cell A without looking at other cells!!!StrmCkr wrote:(5+2) ≠ 7
And this is the most bizarre statement I've seen all year...